Question

The area enclosed by $y=\sqrt{5-x^2}$ and $y=|x-1|$ is

• A. $(\frac{5\pi }{4}-2)$ sq. units
• B. $\left(\frac{5\pi -2}{4}\right)$ sq. units
• C. $(\frac{5\pi }{4}-\frac{1}{2})$ sq. units
• D. $(\frac{\pi }{2}-5)$ sq. units

Answer 1

The correct option is C $(\frac{5\pi }{4}-\frac{1}{2})$ sq. units

Shaded area is the region enclosed by $y=\sqrt{5-x^2}$ and $y=|x-1|$ as shown in figure
The required area = $\underset{-1}{\overset{2}{\int }}\sqrt{5-x^2}-\underset{-1}{\overset{1}{\int }}-(x-1)-\underset{1}{\overset{2}{\int }}(x-1)$
$={[\frac{x\sqrt{5-x^2}}{2}+\frac{5}{2}{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_{-1}^2+{[\frac{x^2}{2}-x]}_{-1}^1-{[\frac{x^2}{2}-x]}_1^2$
$=2+\frac{5}{2}({\sin }^{-1}\frac{2}{\sqrt{5}}+{\sin }^{-1}\frac{1}{\sqrt{5}})-2-\frac{1}{2}$
As we know,${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
$=\frac{5}{2}{\sin }^{-1}[\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}]-\frac{1}{2}$
$=\frac{5}{2}{\sin }^{-1}1-\frac{1}{2}$
$=\frac{5}{2}\times \frac{\pi }{2}-\frac{1}{2}$
$=\frac{5\pi }{4}-\frac{1}{2}$ sq. units